STRUCTURE OF HELIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which could not lead to the correct nuclear structure. For example although there are eight known isotopes of helium physicists influenced by such wrong nuclear theories believe that in nature there exist nine isotopes of helium including the so-called diproton (He-2). In the “Helium-2 WIKIPEDIA” one reads: “There may have been observations of He-2. In 2000, physicists first observed a new type of radioactive decay in which a nucleus emits two protons at once—perhaps a He-2 nucleus.[7][8]The team led by Alfredo Galindo-Uribarri of the Oak Ridge National Laboratory announced that the discovery will help scientists understand the strong nuclear force and provide fresh insights into the creation of elements inside stars.” Under this physics crisis and using the charged UP and DOWN quarks in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” which led to my discovery of the new structure of protons and neutrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002) by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the nuclear binding and nuclear structure by applying the laws of electromagnetism.(See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Although there are eight known isotopes of helium (He) , only helium-3 (He-3) and helium-4 (He-4) are stable. All radioisotopes are short-lived, the longest-lived being He-6 with a half-life of 806.7 milliseconds. The least stable is He-5, with a half-life of 7.6×10−22 seconds. WHY He-4 OF S = 0 AND He-3 OF S = +1/2 ARE STABLE NUCLIDES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single pn bonds of weak horizontal bonds leading to the beta decay. For example in my papers STRUCTURE AND BINDING OF He4 AND He6 , and STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. Also in the structure of He4 we observe two bonds per neutron in which the vertical np bonds are of strong binding energy.For understanding better the structure of He-4 you cna see the following first figure STRUCTURE OF He-6, He-8 He-9 AND He-10 In the following diagrams of He-4 and He-6 you see that the nucleons have opposite spins giving S=0 Of course the He-6 with the same S=0 is unstable because the two extra neutrons outside the He-4 make single horizontal bonds leading to the beta decay. Here you see that the two weak horizontal np bonds exist on the left and on the right side of the stable rectangle of He-4. Now for revealing the structure of He-8 with the same S=0 we see that in the diagram of He-6 the two new extra neutrons like the n5(+1/2) and the n6(-1/2) make single horizontal bonds existing in front of p1 and behind the p2 respectively. So they are not shown in the diagram of He-6. However in the structure of He-9 with S =-1/2 the additional neutron like the n7(-1/2 ) makes a single horizontal np bond existing in front of the p2. So it is not shown in the diagram of He-6 giving S =-1/2. Finally in the He-10 with S =0 the additional neutron like the n8(+1/2) makes with p1 a single horizontal bond existing behind the p1.Therefore all 4 extra neutrons of He-10 which are not shown are of opposite spins giving S=0. ' ' ' Diagrams of the structure of He-4 and He-6 ' '' Helium-4 Helium-6''' ' n2(-1/2)..p2 (- 1/2) n2 (-1/2)..p2(- 1/2)…n4(-1/2)' '' '' ' p1(+1/2)..n1(+1/2) '' ' ' n3(+1/2)… p1(+1/2)..n1(+1/2) '' ' ' '' ''STRUTURE OF He-7 AND He-5 WITH S =-3/2 Using the diagram of He-6 we reveal the structure of He-7. In this case the n3(+1/2) goes from p1 to p2 and changes its spin from +1/2 to -1/2 by making a single horizontal bond existing in front of p2. So it is not shown. Then an additional neutron like the n5(-1/2) makes with p2 a new horizontal bond existing behind it. So it is not shown. Therefore in this structure of He-7 we have two nucleons of positive spins like the p1(+1/2) and the n1(+1/2)while the 5 nucleons like the n2, p2, n3, n4 and n5 have negative spins. Under this condition we get S =+1/2 +1/2 -5/2 = -3/2. Similarly in the structure of He-5 we get the same S =-3/2, because the n1(+1/2) and the n4(-1/2 ) of opposite spins are absent. Category:Fundamental physics concepts